## Pages

`“Life is like riding a bicycle.  To keep your balance you must keep moving.”–Albert Einstein`

## Saturday, August 6, 2011

### AP Physics B - Wave Motion (including sound) - Multiple Choice Practice Questions on Sound

Questions on sound were discussed earlier on this site. You can access all posts related sound by clicking on the label ‘sound’ below this post.
Today we will discuss a few more multiple choice practice questions on sound, relevant to AP Physics B aspirants.
(1) A source producing sound of wave length 0.6 m is moving away from a stationary listener with speed V/6 where V is the speed of sound in air. The wave length of sound heard by the listener is
(a) 0.5 m
(b) 0.54 m
(c) 0.66 m
(d) 0.7m
(e) 0.8 m
You might have come across questions of the above type under Doppler effect.
If the real frequency of the source of sound is n we have
V/n = λ where λ is the wavelength of the sound as measured by the stationary listener when the source is stationary. Therefore we have
V/n = 0.6 ……………..(i)
When the source moves away from the listener at speed V/6 the n sound waves produced per second will occupy a distance V + V/6 so that the wave length as measured by the listener becomes (V + V/6)/n. Therefore we have
(V + V/6)/n = λ1 ……….(ii)
where λ1 is the new wave length.
Dividing Eq (ii) by Eq (i) we obtain
1 + 1/6 = λ1/0.6
Or, 7/6 = λ1/0.6 from which λ1 = 0.7 m.
(2) Two steel wires A and B of the same length are vibrating under the same tension. If the first overtone of wire A has the same frequency as the fundamental note of wire B, the area of cross section of wire A is
(a) twice that of B
(b) three times that of B
(c) four times that of B
(d) half that of B
(e) one third that of B
The frequency of vibration (f) of a string of length and linear density m stretched under tension T is given by
f = (s/2) √(T/m) where s = 1,2,3,4 etc. For the fundamental mode s = 1. For the 2nd mode (or 1st overtone) s = 2. For the 3rd mode (or 2nd overtone) s = 3 and so on.
The linear density is mass per unit length and is equal to aρ where ‘a’ is the area of cross section and ρ is the density of the material of the string.
Since the first overtone of wire A has the same frequency as the fundamental note of wire B, we have
(2/2) √(T/m1) = (1/2) √(T/m2) where m1 is the linear density of A and m2 is the linear density of B.
From mthe above equation it follows that m1 = 4 m2.
Since the wires are of the same material, the area of cross section of A must be 4 times that of B.
[Since the changes are confined to the mode of vibration and the linear density, you should be able to write 2/m1 = 1/m2 to arrive at m1 = 4 m2 and the final answer a1 = 4 a2]
(3) Two sound notes of wave lengths λ1 and λ2 in air produce n beats per second. If λ1 < λ2 the speed of sound in air is
(a) λ1λ2n /(λ2 + λ1)
(b) (λ1 + λ2)n
(c) (λ1λ2 /n )(λ2 + λ1)
(d) λ1λ2n / λ1
(e) λ1λ2n /(λ2 λ1)
The frequency f1 of the sound of wave length λ1 is given by
f1 = v/λ1 where v is the speed of sound in air.
The frequency f2 of the sound of wave length λ2 is given by
f2 = v/λ2
Since λ1 < λ2 we have f1 > f2
Since the number of beats produced per second is n, we have
f1 f2 = n
Therefore, v/λ1 v/λ2 = n
Or, v (11 12) = n
This gives v = λ1λ2n /(λ2 λ1)
(4) A glass tube is open at both ends. The minimum frequency of a tuning fork which vibrates in resonance with the air column in this tube is f. If this tube is held vertically with half its length submerged in water, what will be the minimum frequency of a tuning fork that will vibrate in resonance with the air column in the tube? (a) f/3
(b) f/2
(c) f
(d) 2f
(e) 3f
The air column will vibrate with the minimum frequency in the fundamental mode. When both ends of the glass tube are open, there are anti-nodes at the ends and the adjacent node in the middle so that the length L of the tube is equal to λ/2 where λ is the wave length of the fundamental note. Thus we have
L = λ/2
Or λ = 2L
[Remember that the distance between consecutive anti-nodes (or, consecutive nodes) is equal to λ/2].
When half the length of the glass tube is immersed in water, it becomes a closed pipe of length L/2. In the fundamental mode of vibration of the air column in the tube in this case there is a node at the closed end (water surface) and the adjacent anti-node at the open end as shown in the figure. Therefore we have
L/2 = λ1 /4 where λ1 is the wave length of the resonating sound.
Or λ1 = 2L
In both cases the wave length of the resonating sound is the same. Therefore the resonant frequency is the same as f [Option (c)].
(5) In the above question what will be the minimum frequency of a tuning fork that will vibrate in resonance with the air column in the tube if a quarter of the tube is submerged in water?
(a) 2f/3
(b) 3f/4
(c) f
(d) 2f
(e) 3f
Initially, when both ends are open, the resonating wavelength as shown above is given by
λ = 2L
When a quarter of the glass tube is submerged in water, we have
3L/4 = λ1/4
Therefore λ1 = 3L
From the above equations λ/λ1 = 2/3
Since the frequency is inversely proportional to the wave length we have
λ/λ1 = f1/f
Therefore f1/f = 2/3 from which f1 = 2f/3